SOLUTION: x2 - 8x + 16 >= 0 I got x=4 as the answer but I keep getting told it wrong. I could use help trying to figure out the right answer. A for no x B x <= 4 C -4 <= x <= 4 D

Algebra ->  Inequalities -> SOLUTION: x2 - 8x + 16 >= 0 I got x=4 as the answer but I keep getting told it wrong. I could use help trying to figure out the right answer. A for no x B x <= 4 C -4 <= x <= 4 D       Log On


   

Question 330857: x2 - 8x + 16 >= 0
I got x=4 as the answer but I keep getting told it wrong. I could use help trying to figure out the right answer.

A for no x
B x <= 4
C -4 <= x <= 4
D x = 4
E 2 <= x <= 8
F for all x
Found 2 solutions by Fombitz, Alan3354:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You only have 1/2 of the answer.
x%5E2+-+8x+%2B+16+%3E=+0
%28x-4%29%5E2%3E=0
When you take the square root of both sides you get,
x-4%3E=0 and -%28x-4%29%3E=0
x%3E=4 and x-4%3C=0
x%3E=4 and x%3C=4
which together makes up (-infinity, infinity).
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The function is always greater than or equal to zero as shown in the graph.
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graph%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E2-8x%2B16%29

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve
f(x) = x^2 - 8x + 16 = 0
%28x-4%29%5E2+=0
(x-4) = 0
x = 4
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x=4 is the minimum value of the function, f(x) = 0
For all other values, f(x) > 0
--> For all x, f(x) >= 0
--> All x